# STEP I 2003 4

Write this inequality:

\displaystyle \frac {\sin \theta + 1 - \cos \theta} {\cos \theta} \le 0

So, multiplying by \cos^2 \theta \ne 0:

\displaystyle \cos \theta(\sin \theta + 1 - \cos \theta) \le 0

It must be the case that either:

1. \cos \theta \le 0 and \sin \theta + 1 - \cos \theta \ge 0
2. \cos \theta \ge 0 but \sin \theta + 1 - \cos \theta \le 0

We know that the only solutions of \cos \theta = 0 in this range are \theta = \dfrac \pi 2, \dfrac {3 \pi} 2, giving our first two critical values.

We can start by solving:

\sin \theta - \cos \theta = -1

to find critical points.

This is equivalent to:

\displaystyle \cos \theta - \sin \theta = 1

We aim to write this in the form R \cos (\theta + \alpha), set: \displaystyle R \cos(\theta + \alpha) = R \cos \theta \cos \alpha - R \sin \theta \sin \alpha. Then R = \sqrt 2 and \alpha = \dfrac \pi 4, so we aim to solve:

\displaystyle \sqrt 2 \cos \left(\theta + \frac \pi 4\right) = 1

so that:

\displaystyle \cos \left(\theta + \frac \pi 4\right) = \frac 1 {\sqrt 2}

then either:

\displaystyle \theta + \frac \pi 4 = \frac {\pi} 4 + 2 \pi N

for integer N, ie.

\displaystyle \theta = 2 \pi N

for integer N or:

\displaystyle \theta + \frac \pi 4 = -\frac \pi 4 + 2 \pi N

or:

\displaystyle \theta = -\frac \pi 2 + 2 \pi N

for integer N. The solutions (and therefore critical values) in this range are therefore \displaystyle 0, \frac {3 \pi} 2. So the critical values for this inequality are \displaystyle 0, \dfrac \pi 2, \dfrac {3 \pi} 2. It remains to consider the behaviour of \cos \theta (\sin \theta + 1 - \cos \theta) in the intervals:

• 0 \le \theta \le \dfrac \pi 2
• \dfrac \pi 2 \le \theta \le \dfrac {3 \pi} 2
• \dfrac {3 \pi} 2 \le \theta \le 2 \pi

In the first interval, we have \cos \theta \le 0, in the second we have \cos \theta \ge 0 and in the third we have \cos \theta \le 0. It just remains to determine the sign of \sin \theta + 1 - \cos \theta.

In the first case, we check if \sin \theta + 1 - \cos \theta \ge 0, ie. if \displaystyle \cos \left(\theta + \frac \pi 4\right) \ge \frac 1 {\sqrt 2}. Note that when \theta = 0, we do have \displaystyle \cos \left(\theta + \frac \pi 4\right) = \frac 1 {\sqrt 2}, but since cosine is decreasing for 0 \le \theta \le \pi, we cannot have \displaystyle \cos \left(\theta + \frac \pi 4\right) \ge \frac 1 {\sqrt 2} at any other point. So the inequality is satisfied in this case only if \theta = 0.

In the second case, we check if \sin \theta + 1 - \cos \theta \le 0, ie. if \displaystyle \cos \left(\theta + \frac \pi 4\right) \le \frac 1 {\sqrt 2}. At \theta = \dfrac \pi 2, we have \displaystyle \cos \left(\theta + \frac \pi 4\right) = -\frac 1 {\sqrt 2}. At this point, the LHS is decreasing with increasing \theta. It continues to decrease until \displaystyle \theta = \pi - \frac \pi 4 = \dfrac {3 \pi} 4, then increases again up until \displaystyle \theta = 2 \pi - \frac \pi 4 = \frac {7 \pi} 4. Since at \theta = \dfrac {3 \pi} 2 we have \displaystyle \cos \left(\theta + \frac \pi 4\right) = \frac 1 {\sqrt 2}, this means this inequality is satisfied for \displaystyle \frac \pi 2 \le \theta \le \frac {3 \pi} 2. However since we insisted that \cos \theta \ne 0, we cannot have \displaystyle \theta = \frac \pi 2 or \displaystyle \theta = \frac {3 \pi} 2.

A similar analysis to the above with flipped signs reveals that the inequality also holds for \displaystyle \frac {3 \pi} 2 < \theta < \pi.

So the original inequality has solution \displaystyle \{0\} \cup \left\{\theta \ne \frac {3 \pi} 2 \mid \frac \pi 2 < \theta < 2 \pi\right\}

Alternative solution (realised this the LHS looked familiar)

Note that:

\begin{align*}\tan \left(\frac \theta 2 + \frac \pi 4\right) & = \frac {\tan \frac \theta 2 + \tan \frac \pi 4} {1 - \tan \frac \theta 2 \tan \frac \pi 4} \\ & = \frac {\tan \frac \theta 2 + 1} {1 - \tan \frac \theta 2} \\ & = \frac {\cos \frac \theta 2 + \sin \frac \theta 2} {\cos \frac \theta 2 - \sin \frac \theta 2} \\ & = \frac {\left(\cos \frac \theta 2 + \sin \frac \theta 2\right)^2} {\left(\cos \frac \theta 2 - \sin \frac \theta 2\right)\left(\cos \frac \theta 2 + \sin \frac \theta 2\right)} \\ & = \frac {\cos^2 \frac \theta 2 + \sin^2 \frac \theta 2 + 2 \sin \frac \theta 2 \cos \frac \theta 2} {\cos^2 \frac \theta 2 - \sin^2 \frac \theta 2} \\ & = \frac {1 + \sin \theta} {\cos \theta}\end{align*}

The inequality is the question is then equivalent to:

\displaystyle \displaystyle \tan \left(\frac \theta 2 + \frac \pi 4\right) \le 1

Note that \tan \left(0 + \frac \pi 4\right) = 1, and then \displaystyle \tan \left(\frac \theta 2 + \frac \pi 4\right) is increasing for increasing \theta (so there can be no further solutions in the interval 0 < \theta < \dfrac \pi 2) until the discontinuity at \theta = \dfrac \pi 2, at which point it will continue increasing until the next discontinuity at \theta = \dfrac {5 \pi} 2, but this is outside the interval we’re interested in, we’re only interested in what happens up to \theta = 2 \pi. Note that in fact \displaystyle \tan \left(\frac {2 \pi} 2 + \frac \pi 4\right) = \tan \dfrac {5 \pi} 4 = 1, so \displaystyle \tan \left(\frac \theta 2 + \frac \pi 4\right) \le 1 for \displaystyle \dfrac \pi 2 < \theta < 2 \pi. Hence our conclusion with the last method.

With this trick it’s one of the shortest STEP questions.

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